Metamath Proof Explorer


Theorem scottab

Description: Value of the Scott operation at a class abstraction. (Contributed by Rohan Ridenour, 14-Aug-2023)

Ref Expression
Hypothesis scottab.1 x = y φ ψ
Assertion scottab Scott x | φ = x | φ y ψ rank x rank y

Proof

Step Hyp Ref Expression
1 scottab.1 x = y φ ψ
2 nfv x ψ
3 2 1 scottabf Scott x | φ = x | φ y ψ rank x rank y